Iron burns in oxygen to form iron oxide Fe3O4. Determine the mass of iron scale that is formed during the combustion of 47.6 g of iron. How many liters of oxygen is required for the iron to react completely.
Let’s implement the solution:
1. We write down the process according to the condition of the problem:
m = 47.6 g. X l. -? X g -?
3Fe + 2O2 = Fe3O4 – compounds, scale is released;
2. Calculation by formulas:
M (Fe) = 55.8 g / mol;
M (O2) = 32 g / mol;
M (Fe3O4) = 231.4 g / mol;
Y (Fe) = m / M = 47.6 / 55.8 = 0.85 mol. (substance in excess)
0.85 mol (Fe) – X mol (O2);
-3 mol – 2 mol from here, X mol (O2) = 0.85 * 2/3 = 0.57 mol, (the substance is in short supply).
Calculations are made for the substance in deficiency.
0.57 mol (O2) – X mol (Fe3O4);
-2 mol -1 mol from here, X mol (Fe3O4) = 0.57 * 1/2 = 0.28 mol.
4. Find the volume of O2, the mass of scale:
V (O2) = 0.57 * 22.4 = 12.77 L;
m (Fe3O4) = Y * M = 0.28 * 231.4 = 64.79 g.
Answer: for oxidation, oxygen with a volume of 12.77 liters is required, and iron oxide with a mass of 64.79 g is obtained.
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